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Relation between coefficients and zeros of a quadratic polynomial

  1. May 20, 2012 #1
    1. The problem statement, all variables and given/known data

    For any quadratic polynomial ax2+bx+c having zeros β and α
    Prove that β + α = -b/a and αβ = c/a.

    2. Relevant equations



    3. The attempt at a solution

    I have found a method myself to prove α+ β = -b/a. However, I could not prove αβ = c/a.

    It goes like this.

    If α and β are the zeros of the given polynomial.

    a(α)2+b(α) + c = 0 .................... (i)

    Also,

    a(β)2+b(β) + c = 0 ....................(ii)

    Comparing (i) and (ii)

    a(α)2+b(α) + c = a(β)2+b(β) + c

    => aα2-aβ2 = bβ - bα
    =>a(α22) = -b(α-β)
    =>α22/α-β = -b/a
    => α+β = -b/a

    Please help me prove αβ = c/a using the same method.
     
    Last edited: May 20, 2012
  2. jcsd
  3. May 20, 2012 #2

    SammyS

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    Use the quadratic formula for the two zeros of the quadratic polynomial.
     
  4. May 20, 2012 #3

    Hurkyl

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    Do you have reason to think the same method will work?

    It would be easier to just write down the factorization of the polynomial, I think.
     
  5. May 20, 2012 #4

    NascentOxygen

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    Given: aα²+bα + c = 0 .................... (i)
    it factorizes: aα(α + b/a) + c = 0

    Substitute for the bold α.
     
  6. May 20, 2012 #5
    Thank you very very very very much. I can't explain how happy I am to get the answer. Thanks again.
     
  7. May 20, 2012 #6

    SammyS

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    Yes, that's correct supposing that physics kiddy knows:
    If α and β are the solutions to [itex]\text{a}x^2+bx+c=0\,,[/itex] then [itex]\text{a}x^2+bx+c=\text{a}(x-\alpha)(x-\beta)\ .[/itex]​

    Of course, I agree that is a very handy thing to know!
     
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